# Limit Formula

Last Updated : 27 Feb, 2024

If a given function y = f(x) is not defined at the point x = a. but the function can be defined at neighborhood values of x = a. That’s the limiting behavior of a function. limit helps us to determine the values of a function at x = a, but not the exact value, it’s an approaching value of x at a. When x approaches a and we can write as limxâ‡¢a f(x). The approaching and exact value has a very very small difference between them i.e. if the exact point is 2 then the approaching value is 1.9999999… so on. This difference can’t affect any major.

Most students believe that a limit should be a finite number. But it is quite possible that f(x) had an infinite limit as xâ‡¢a. i.e.  limxâ‡¢a f(x) = âˆž.

### Limit Formulae

Trigonometric limits: To evaluate trigonometric limits, we have to reduce the terms of the function into simpler terms or into terms of sinÎ¸ and cosÎ¸.

• limx â‡¢ 0  [Tex]\frac{sinx}{x} [/Tex] = limx â‡¢ 0 [Tex]\frac{x}{sinx} [/Tex] = 1
• limx â‡¢ 0 tanx/x = limx â‡¢ 0 x/tanx =1

As we considered our first one,

limx â‡¢ 0 sinx/x =1

Using L-Hospital

limx â‡¢ 0 cosx/1

limx â‡¢ 0 cos(0)/1 = 1/1 =1

If the function gives an indeterminate form by putting limits, Then use the l-hospital rule.

Indeterminate  Form

0/0, âˆž/âˆž, âˆž-âˆž, âˆž/0, 0âˆž, âˆž0 , 00, âˆžâˆž

L-hospital Rule

If we get the indeterminate form, then we differentiate the numerator and denominator separately until we get a finite value. Remember we would differentiate the numerator and denominator the same number of times. Similarly for all trigonometric function,

• limx â‡¢ 0 sin-1x/x = limx â‡¢ 0 x/sin-1x = 1

limx â‡¢ 0 sin-1x/x =1

limx â‡¢ 0 1/âˆš1+x2  [Using L-Hospital]

= 1/âˆš(1 + (0)2) = 1

• limxâ‡¢0[Tex] \frac{tan^{-1}x}{x}  [/Tex]=1
• limx â‡¢ a  sin xo/x = Ï€/180
• limxâ‡¢0 cosx = 1
• limx â‡¢ a  sin(x-a) / (x-a) =1

limx â‡¢ a sin(x – a) / (x – a)

=1

limx â‡¢ a cos(x – a)/1

= limx â‡¢ a cos(a – a) = cos(0) =1

• limxâ‡¢âˆž sinx/x = 0
• limxâ‡¢âˆž cosx/x = 0
• limxâ‡¢âˆž sin(1/x) / (1/x) =0

limx â‡¢ âˆž sin(1/x)/(1/x) = 0

Let 1/x = h

So, limits changes to 0

Because 1\âˆž = 0

limh â‡¢ 0 sinh/h

As we see before, If limx â‡¢ 0 sinx/x = 1

So, limh â‡¢ 0 sinh/h = 1

Exponential limits

• limx â‡¢ 0  ex – 1 /x = 1
• limx â‡¢ 0  ax – 1 /x = logea
• limx â‡¢ 0 eÎ»x – 1 /x = Î»

Same here, we get our desired result by using L-hospital rule.

Alternate method: Using expansion

ex = 1 + X + X2/2! + X3/3! + X4/4!+ … âˆž

limx â‡¢ 0  ex – 1 /x = 1

limx â‡¢ 0 (1 + X + X2/2!+ —) -1 /x

limx â‡¢ 0 (X + X2/2! + —)/x

limx â‡¢ 0 1 + X + X2/2!+—

limx â‡¢ 0 1 + 0 + 0 + 0 + 0— = 1

Logarithmic limits

• limx â‡¢ 0 log(1 + x) /x = 1
• limx â‡¢ e logex = 1
• limx â‡¢ 0 loge(1 – x) /x  = -1
• limx â‡¢ 0 loga(1 + x) /x = logae

Simply proved by using L-hospital and expansion method.

Some Important expansion

[Tex]e^x = 1+\frac{x}{1!} +\frac{x^2}{2!} + \frac{x^3}{3!} +— [/Tex]
[Tex]a^x = 1+ xloga + \frac{x^2(loga)^2}{2!} + \frac{x^3(loga)^3}{3!} +— [/Tex]
[Tex]log(1+x) = x- \frac{x^2}{2} + \frac{x^3}{3} -\frac{x^4}{4} +— [/Tex]
[Tex]log(1-x) = -x- \frac{x^2}{2} – \frac{x^3}{3} -\frac{x^4}{4} +— [/Tex]
[Tex](1+x)n = 1 + nx + \frac{n(n-1)x^2}{2!} +— [/Tex]
[Tex]sinx = x- \frac{x^3}{3!} + \frac{x^5}{5!} +— [/Tex]
[Tex]cosx = 1- \frac{x^2}{2!} + \frac{x^4}{4!} +— [/Tex]
[Tex]tanx = x + \frac{x^3}{3} + \frac{2x^5}{15} +— [/Tex]
[Tex]sin^{-1}x = x + \frac{x^3}{3!} + \frac{9x^5}{5!} +— [/Tex]
[Tex]cos^{-1}x = x – \frac{x^3}{6} + — [/Tex]
[Tex]tan^{-1}x= x – \frac{x^3}{3} + \frac{x^5}{5} +— [/Tex]
[Tex]sinhx =  x+ \frac{x^3}{3!} + \frac{x^5}{5!} +—        [/Tex] (Here, sinhx is a hyperbolic function)
[Tex]coshx = 1+ \frac{x^2}{2!} + \frac{x^4}{4!} +— [/Tex]
[Tex]tanhx = x – \frac{x^3}{3} + \frac{2x^5}{15} +— [/Tex]
[Tex](1+x)\frac{1}{x} = e^{1- \frac{x}{2} + \frac{11×4}{24} } +— [/Tex]

### Important Results :

1. limxâ‡¢0 [Tex](1+x)^{\frac{1}{x}}  [/Tex]= e
2. limxâ‡¢0 [Tex]\left( 1+\frac{1}{x} \right)^x [/Tex] = e
3. limxâ‡¢0 [Tex]\frac{e^x-1}{x} =1 [/Tex]
4. limxâ‡¢0[Tex]\frac{a^x-1}{x} [/Tex] = logea
5. limxâ‡¢0[Tex] \frac{1-cosmx}{x^2} [/Tex] = m2/2
6. limxâ‡¢0 [Tex]\frac{1-cosmx}{1-cosnx} = \frac{m^2}{n^2} [/Tex]

### Shortcut :

1. limxâ‡¢0 [Tex]\left( 1+ \frac{a}{b}x \right)^\frac{c}{dx} = e^\frac{ac}{bd} [/Tex]
2. limxâ‡¢0[Tex] \left( 1+ \frac{a}{bx} \right)^\frac{cx}{d} = e^\frac{ac}{bd} [/Tex]
3. limxâ‡¢a f(x)g(x) = e^{limx\to a\left[ f(x)-1 \right].g(x)}

### Sample problem

Question 1: Solve, limxâ‡¢0  (x – sinx ) /(1 – cosx).

Solution:

Using L-hospital,

limx â‡¢ 0 (1 – cosx) / (sinx)

limx â‡¢ 0 sinx / cosx = sin(0) / cos(0) = 0/1 = 0

Question 2: Solve, limx â‡¢ 0 (e2x -1) / sin4x.

Solution:

Using L-hospital

limx â‡¢ 0 (2)(e2x) / cos4x

limx â‡¢ 0  2(e0) / cos4(0) = 2/1= 2

Question 3: Solve, limx â‡¢ 0 (1 – cosx) / x2

Solution:

Using L-hospital

limx â‡¢ 0 sinx /2x = 1/2 {sinx/x = 1}

Question 4: Solve, limx â‡¢ âˆž [Tex]\frac{x+sinx}{x} [/Tex]

Solution:

limx â‡¢ âˆž (1 + [Tex]\frac{sinx}{x}        [/Tex])

1 + limx â‡¢ âˆž[Tex] \frac{sinx}{x} [/Tex]

As we know, x = âˆž

So 1/x = 0

[Tex]1 + lim\frac{1}{x}â‡¢âˆž \frac{sin\frac{1}{x}}{x} [/Tex]

1 + 0 = 0

Question 5: Solve, limx â‡¢ Ï€/2 (tanx)cosx

Solution:

let Y = limx â‡¢ Ï€/2  (tanx)cosx

Taking loge both sides,

logeY = limx â‡¢ Ï€/2  loge(tanx)cosx

logeY = limx â‡¢ Ï€/2  cosx loge(tanx)

logey = limx â‡¢ Ï€/2 loge(tanx)/secx

Using l-hospital,

logey = limx â‡¢ Ï€/2 cosx /sin2x = 0

Now, taking exponent on both sides,

Y = limx â‡¢ Ï€/2 e0

Y = limx â‡¢ Ï€/2  (tanx)cosx = 1

Question 6: limx â‡¢ 0  [Tex]\frac {e^x -(1+x+ \frac {x^2}{2})}{ x^3} [/Tex]

Solution:

limxâ‡¢0 \frac{1+\frac{x}{1!} + \frac{x2}{2!} + \frac{x3}{3!} – ( 1+ x+ \frac{x2}{2!} ) }{x3}

limxâ‡¢0 [Tex]\frac{\frac{x3}{3!}}{x3}  [/Tex]= 1/3! =1/6

Question 7: Solve, lima â‡¢ 0  [Tex]\frac{x^a-1}{a} [/Tex]

Solution:

Using l-hospital (Differentiating numerator and denominator w.r.t a)

lima â‡¢ 0  xalogx = logx

Question 8: Solve, limx â‡¢ 0 [Tex]\frac{x^2+x-sinx}{x^2} [/Tex]

Solution:

limx â‡¢ 0 [Tex]\frac{x^2+x-(x-x^3/3!)}{x^2} [/Tex]

limx â‡¢ 0 1 + x/3! = 1

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